http://acm.hdu.edu.cn/showproblem.php?pid=5898
Problem Description
For a number,if the length of continuous odd digits is even and the length of continuous even digits is odd,we call it odd-even number.Now we want to know the amount of odd-even number between L,R(1<=L<=R<= 9*10^18).
Input
First line a t,then t cases.every line contains two integers L and R.
Output
Print the output for each case on one line in the format as shown below.
Sample Input
2
1 100
110 220
Sample Output
Case #1: 29
Case #2: 36
1 #include <algorithm> 2 #include <iostream> 3 #include <cstring> 4 #include <cstdlib> 5 #include <cstdio> 6 #include <vector> 7 #include <ctime> 8 #include <queue> 9 #include <list> 10 #include <set> 11 #include <map> 12 using namespace std; 13 #define INF 0x3f3f3f3f 14 typedef long long LL; 15 16 int bit[20]; 17 LL dp[20][20][2];dp[len][sum][tag]//前len 位，到当前为为止有num 个tag, tag=0 表示偶数，1为奇数 18 LL dfs(int len, int zer, int num, int tag, int flag) 19 { 20 if(len < 1) 21 return (num + tag) % 2 || (zer==0); 22 if(flag && dp[len][num][tag] != -1) 23 return dp[len][num][tag]; 24 int te = flag ? 9 : bit[len]; 25 LL sum = 0; 26 for(int i = 0; i <= te; i++) 27 { 28 if(i % 2 == 0){ 29 if(zer == 0 && i == 0) 30 sum += dfs(len-1, 0, 0, 0, flag || i < te); 31 else if(tag == 0) 32 sum += dfs(len-1, 1, num+1, tag, flag || i < te); 33 else if(tag == 1 && num % 2 == 0) 34 sum += dfs(len-1, 1, 1, 0, flag || i < te); 35 } 36 if(i % 2 == 1){ 37 if(tag == 1) 38 sum += dfs(len-1, 1, num+1, tag, flag || i < te); 39 else if(zer==0 || (tag == 0 && num % 2 == 1)) 40 sum += dfs(len-1, 1, 1, 1, flag || i < te); 41 } 42 } 43 if(flag) 44 dp[len][num][tag] = sum; 45 return sum; 46 } 47 LL solve(LL n) 48 { 49 memset(bit, 0, sizeof(bit)); 50 int len = 0; 51 while(n) 52 { 53 bit[++len] = n % 10; 54 n /= 10; 55 } 56 return dfs(len, 0, 0, 0, 0); 57 } 58 int main() 59 { 60 LL l, r; 61 int t; 62 scanf("%d", &t); 63 memset(dp, -1, sizeof(dp)); 64 for(int ca = 1; ca <= t; ca++) 65 { 66 scanf("%lld%lld", &l, &r); 67 printf("Case #%d: %lld ", ca, solve(r)-solve(l-1)); 68 } 69 return 0; 70 }